Consider $\mathbb{F}_2^{2\times2}$, the $2\times2$-matrices over the finite field $\mathbb{F}_2$. It seems to me (by trial and intuition, if I'm being honest), that there should be no matrix (besides $\mathbf{1}, \mathbf{0}$) that would commute with every invertible matrix in $\mathbb{F}_2^{2\times2}$.
Note that I am not requiring that this matrix commute with all other matrices in $\mathbb{F}_2^{2\times2}$, only with the invertible ones. For one, we do know that the group of invertible matrices in $\mathbb{F}_2^{2\times2}$ has trivial center, so I would only need to check singular matrices.
I tried to prove this by brute-force calculation, but since that is rather tedious, I would be interested to know a more analytical approach to this problem (or if I'm mistaken entirely).
Best Answer
Fix $B$ non-invertible. Let $\text{im }B = \text{span } v$ for some nonzero $v$.
Then if $(v,w)$ is basis for $\Bbb F_2$ define $A$ by $v \mapsto w$, $w \mapsto v$.
Clearly $\text{im }BA = \text{im }B = \text{span } v \ne \text{span } w = \text{im }AB$.
So we can always find an invertible $A$ that doesn't commute with a given non-invertible $B \ne 0$.